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Proof step 12 from proof addobj-in-02 (conclusion)

⊦ r ≠ r0 ∨ ¬ sk ∈ sks ∧ sk ≠ _type → (r -- sk ∈ addobj(r0, class, sks, st) ↔ r -- sk ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
r ≠ r0 ∨ ¬ sk ∈ sks ∧ sk ≠ _type ⊦ r -- sk ∈ addobj(r0, class, sks, st) ↔ r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]